The number of real roots of the equation $x |x - 2| + 3|x - 3| + 1 = 0$ is :

<p>$$ \begin{aligned} & \text { (I) } x<2 \\ & -x^2+2 x-3 x+9+1=0 \\ & \Rightarrow x^2+x-10=0 \\ & \Rightarrow x=\frac{-1+\sqrt{41}}{2}, \frac{-1-\sqrt{41}}{2} \\ & \quad \quad \qquad \quad \times \qquad \qquad \sqrt{ } \end{aligned}$$</p> <p>$$\begin{aligned} & \text { (II) } 2 \leq x<3 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}-3 \mathrm{x}+9+1=0 \\ & \Rightarrow \mathrm{x}^2-5 \mathrm{x}+10=0 \\ & \mathrm{D}<0 \\ & \text { (III) } \mathrm{x} \geq 3 \\ & \mathrm{x}^2-2 \mathrm{x}+3 \mathrm{x}-9+2=0 \\ & \Rightarrow \mathrm{x}^2+\mathrm{x}-8=0 \end{aligned}$$</p> <p>$\mathrm{x}=\frac{-1+\sqrt{32}}{2}, \frac{-1-\sqrt{32}}{2}$</p> <p>$\quad \quad \times \qquad \qquad \times$</p> <p>1 real roots</p>